
Hence, using (), (REF ), (E.4), (E.13), and (E.14)
[1]}, one infers that
\(&\partial _{t_r} \Big (\sum _{j=0}^n \int _{Q_0}^{\hat{\mu }_j}\underline{\omega }\Big )=\sum _{j=0}^n \mu _{j,t_r} \sum _{k=1}^n\underline{c}(k)\frac{a~\mu _j^{k-1}}{y(\hat{\mu }_j)}\nonumber \\&=4a\sum _{j=0}^n \sum _{k=1}^n \underline{c}(k)\frac{\mu _j^{k-1}}{\prod _{\scriptstyle l=0 \atop \scriptstyle l \ne j}^n (\mu _j-\mu _l)}\frac{\widetilde{F}_{r}(\mu _j)}{\mu _j}\nonumber \\&=4a\sum _{j=0}^n \sum _{k=1}^n \underline{c}(k)\frac{\mu _j^{k-1}}{\prod _{\scriptstyle l=0 \atop \scriptstyle l \ne j}^n (\mu _j-\mu _l)}\nonumber \\&\times \Big (-\sum _{m=0}^{(r+1) \wedge (n+1)} \tilde{d}_{r+1,m}(\underline{E})\Psi _{m}(\bar{\mu })\frac{\Phi _{n}^{(j)}(\bar{\mu })}{\Psi _{n+1}(\bar{\mu })}+\sum _{m=1}^{(r+1) \wedge (n+1)} \tilde{d}_{r+1,m}(\underline{E})\Phi _{m-1}^{(j)}(\bar{\mu }) \Big )\nonumber \\&=-4a\sum _{m=0}^{(r+1) \wedge (n+1)} \tilde{d}_{r+1,m}(\underline{E})\frac{\Psi _{m}(\bar{\mu })}{\Psi _{n+1}(\bar{\mu })}\sum _{k=1}^n \sum _{j=0}^n \underline{c}(k)(U_{n+1}(\bar{\mu }))_{k,j}(U_{n+1}(\bar{\mu }))_{j,1}^{-1}\nonumber \\&~~~~~+4a \sum _{m=1}^{(r+1) \wedge (n+1)} \tilde{d}_{r+1,m}(\underline{E})\sum _{k=1}^n \sum _{j=0}^n \underline{c}(k)(U_{n+1}(\bar{\mu }))_{k,j}(U_{n+1}(\bar{\mu }))_{j,n-m+2}^{-1}\nonumber \\&=-\frac{4a}{\Psi _{n+1}(\bar{\mu })}\sum _{m=0}^{(r+1) \wedge (n+1)} \tilde{d}_{r+1,m}(\underline{E})\Psi _{m}(\bar{\mu }) \underline{c}(1)+4a\sum _{m=1}^{(r+1) \wedge (n+1)} \tilde{d}_{r+1,m}(\underline{E})\underline{c}(n-m+2)\nonumber \\&=-\frac{4a}{\Psi _{n+1}(\bar{\mu })}\sum _{m=0}^{(r+1) \wedge (n+1)} \tilde{d}_{r+1,m}(\underline{E})\Psi _{m}(\bar{\mu }) \underline{c}(1)+4a\sum _{m=1\vee (n+1-r) }^{ n+1} \tilde{d}_{r+1,n+2-m}(\underline{E})\underline{c}(m), \) 
